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## Main Question or Discussion Point

I am kind of new to this eigenvalue, eigenfunction and operator things, but i have come across this quote many times:

\begin{align}

\langle W \rangle &= \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x

\end{align}

From which it follows that ##\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p##.

I know how to derive relation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## for which they state that:

First i need some explaination on how do we know this? All i know about operator ##\hat{H}## so far is this equation where ##\langle W \rangle## is an energy expected value:##\psi## is the eigenfunction of an operator ##\hat{H}## with eigenvalue ##W##.

\begin{align}

\langle W \rangle &= \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x

\end{align}

From which it follows that ##\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p##.

**Aditional question:**I know how to derive relation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## for which they state that:

I also know how to derive relation ##\hat{H}\hat{a}^\dagger = (W + \hbar \omega)\hat{a}^\dagger \psi## for which they state that:##\hat{a} \psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W-\hbar \omega)##.

How do we know this?##\hat{a}^\dagger \psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W+\hbar \omega)##.